Which of the following is a solution of the equation x - 7 = - 12?
Which of the following is a solution of the equation x - 7 = - 12? => - 5 or 5 or - 19 or 19..
Purification of Colloidal Sols
Purification of Colloidal Sols - The colloidal solutions prepared by various methods usually contain electrolytes and other soluble substances as impurities. These impurities if not removed can destabilize the sols. Impurities are removed by the following met..
Which of the following is the same as 12 - 7?
Which of the following is the same as 12 - 7? => 10 - 6 or 6 + 7 or 4 × 2 or 10 ÷ 2..
Solution:
In 100g of the compound, masses of elements are as follows: In 100 grams of compound, 71.8 g are antimony and 28.2% sulphur. Empirical formula is Sb 2 S ..
In 100g of the compound, masses of elements are as follows: In 100 grams of compound, 71.8 g are antimony and 28.2% sulphur. Empirical formula is Sb 2 S ..Solution
Potassium dichromate = K 2 Cr 2 O 7 GMM = (2 x 39) + (2 x 52) + (7 x 16) = 78 + 104 + 112 g = 294 ..
Potassium dichromate = K 2 Cr 2 O 7 GMM = (2 x 39) + (2 x 52) + (7 x 16) = 78 + 104 + 112 g = 294 ..Solution:
Cl = 35.5 1 mole of a substance = 22.4L 1 Mole of a substance = 1 GMM 1 GMM of Cl 2 = 71 g 71 g of Cl 2 = 22.4 L 7.1 g of Cl 2 = ? 7.1 g of Cl 2 will occupy a volume of 2.24 lite..
Cl = 35.5 1 mole of a substance = 22.4L 1 Mole of a substance = 1 GMM 1 GMM of Cl 2 = 71 g 71 g of Cl 2 = 22.4 L 7.1 g of Cl 2 = ? 7.1 g of Cl 2 will occupy a volume of 2.24 lite..Solution
Atomic number Z = 12, Mass number A = 25 Number of protons = Z = 12 Number of neutrons = A number of protons = 25 - 12 = 13 2. Find (i) the total number of neutrons and (ii) total mass of neutrons in 7 mg of 1 4 C. Assume: mass of neutron = mass of hyd..
Atomic number Z = 12, Mass number A = 25 Number of protons = Z = 12 Number of neutrons = A number of protons = 25 - 12 = 13 2. Find (i) the total number of neutrons and (ii) total mass of neutrons in 7 mg of 1 4 C. Assume: mass of neutron = mass of hyd..Solution
In the first compound hydrogen = 5.93% Oxygen = (100 -5.93) = 94.07% In the second compound Hydrogen = 11.2% Oxygen = (100 -11.2) = 88.88% In the first compound the number of parts of oxygen that combine with one part by mass of hydrogen In the second compound the number of parts by mass of oxyg..
In the first compound hydrogen = 5.93% Oxygen = (100 -5.93) = 94.07% In the second compound Hydrogen = 11.2% Oxygen = (100 -11.2) = 88.88% In the first compound the number of parts of oxygen that combine with one part by mass of hydrogen In the second compound the number of parts by mass of oxyg..Solution
Nitrobenzene is reduced to aniline as follows: 123 g 6F (6 x 96,5000) So, if the current efficiency is 100%, then 6 x 96500 C of electricity would be required to reduce 123 g of nitrobenzene. Then, Quantity of electricity (at 50% efficiency) required to reduce 12.3 g of n..
Nitrobenzene is reduced to aniline as follows: 123 g 6F (6 x 96,5000) So, if the current efficiency is 100%, then 6 x 96500 C of electricity would be required to reduce 123 g of nitrobenzene. Then, Quantity of electricity (at 50% efficiency) required to reduce 12.3 g of n..Solution
For the first compound, mass % of C = 42.9 Mass % of O = 57.1 Thus, 42.9 g of C reacts with 57.1 g of oxygen =1.33 g of oxygen For the second compound, mass % of C = 27.3 Mass % of O = 72.7 Thus, 27.3 g of C reacts with 72.7 g of oxygen = 2.66 g of oxygen The ratio of oxy..
For the first compound, mass % of C = 42.9 Mass % of O = 57.1 Thus, 42.9 g of C reacts with 57.1 g of oxygen =1.33 g of oxygen For the second compound, mass % of C = 27.3 Mass % of O = 72.7 Thus, 27.3 g of C reacts with 72.7 g of oxygen = 2.66 g of oxygen The ratio of oxy..See what our Users say :
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